Question

The hybridisation and geometry of $B$ and $N$ in $[H_{3}B\leftarrow N{H}_3]$ are, respectively

• A. $s{p}^3$ , tetrahedral and $s{p}^3$ pyramidal
• B. $s{p}^3$ , pyramidal and $s{p}^3$ tetrahedral
• C. $s{p}^3$ , pyramidal and $s{p}^3$ pyramidal
• D. $s{p}^3$ , tetrahedral and $s{p}^3$ tetrahedral

Answer 1

Answer: (C)

Since, $B{F}_3$ is electron deficient and needs two electrons to complete its octet. It acts as a Lewis acid and reacts with $NH$, which is a Lewis base due to the presence of lone pair of $e$ - on $N$-atom. Acceptance of a pair of electrons from $N{H}_3$ completes the octet of $B$.

Ceptral atom: Boron
Hybridisation: $s{p}^2$
Geometry: tetrahedral
In $B{F}_3\to N{H}_3$. Both $N$ and $B$ contains atoms with
$s{p}^3$ -hybridisation with tetrahedral geometry.